The effects of gravity on pressure can be expressed as: dP = gdz where is densit

Question

The effects of gravity on pressure can be expressed as:

dP = gdz  

where is density and g is the acceleration due to Earth’s gravitational field and z is the distance from sea-level (altitude).

a. Show how we can solve this equation for an ideal gas to yield the change in pressure from an initial value Pi to a final value Pf at the altitudes i z and f z respectively. The expression you get relates the distribution of gas molecules in the atmosphere as a function of their molar mass (M), height, temperature and the acceleration due to gravity and is known as the barometric distribution law.

b. Consider an atmosphere that has a composition of 2 0.6 Nx = and 2 0.4 Ox = and that T=300 K. Near sea level, the total pressure is 1.00 bar. Calculate the mole fractions of the two components at a height of 50.0 km. Assume that the temperature is the same at a height of 50 km.

Explanation / Answer

dP = gdz    (1)

for ideal gas PV= nRT

or PV= ( mass/molar mass)*RT

P* Molar mass = (Mass/Volume)*RT

P* Molar masss = *RT

P*Molar mass/RT=

dP = -P*(Molar mass/RT)g*dZ

or Eq.1 becomes

dP/P= -(M/RT)gdZ, M = Molar mass of air

when this expression is integrated

ln(P2/P1)= -(Mg/RT)*(Z2-Z1), P2 = pressure at height Z2 and P1= pressure at height Z1.

At 50m height, Z2-Z1= 50m, g= 9.8 m/sec2, M= 0.0289 Kg/mole. R= 8.314 J/mole.K

Ln(P2/P1)= -0.0289*50*9.8/(8.314*300)

P2/P1= 0.994

Or P2= 0.994 bar

Hence composition at 50m height =

Nx=20.6*0.994 =20.47 and Ox= 20.4*0.994= 20.3

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